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Learning objectives
Core idea
Key formulae
Worked examples
Explore the graph
Common mistakes
Exam tip
Guided practice
Practice questions
Challenge
Summary
Quick quiz
Same topic, different board
Past paper questions

Learn7367Paper 2 · Compulsory contentHyperbolic functionsHyperbolic functions and their inverses

Hyperbolic functions · Subtopic 1 of 1

Hyperbolic Functions

$\sinh x=\frac{e^x-e^{-x}}2$ , $\cosh x=\frac{e^x+e^{-x}}2$ , $\tanh x=\frac{\sinh x}{\cosh x}$ . They obey identities like trig ones ( $\cosh^2x-\sinh^2x=1$ ), and their inverses are logarithms, e.g. $\operatorname{arsinh}x=\ln\left(x+\sqrt{x^2+1}\right)$ .

Cambridge 9231 Paper 2Edexcel YFM01 WFM03Edexcel 9FM0 Paper 1Edexcel 9FM0 Paper 2AQA 7367 Paper 1AQA 7367 Paper 2≈ 45 minRevision 1 · updated 2026-10-27

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Section 1 of 14

Learning objectives

Define and sketch $\sinh$ , $\cosh$ , $\tanh$ .
Prove and use hyperbolic identities (Osborn's rule).
Express inverse hyperbolic functions as logarithms.
Solve equations involving hyperbolic functions.

Pause and think

A chain hanging between two posts makes a curve. Is it a parabola?

Section 2 of 14

Core idea

Start hereThe shape of a hanging chain

Hang a necklace from two fingers. Galileo thought the curve was a parabola; he was wrong. It's a catenary, $y=\cosh x$ , built from exponentials. Its relatives $\sinh$ and $\tanh$ behave uncannily like sine and tangent, but for a hyperbola instead of a circle.

The big idea

$\sinh x=\frac{e^x-e^{-x}}2$ , $\cosh x=\frac{e^x+e^{-x}}2$ , $\tanh x=\frac{\sinh x}{\cosh x}$ . Identities: $\cosh^2x-\sinh^2x=1$ (Osborn's rule). Inverses are logarithms.

1. Definitions and identities

\cosh^2x-\sinh^2x=1

$\cosh^2x=\tfrac14(e^{2x}+2+e^{-2x})$
Square the definition.
$\sinh^2x=\tfrac14(e^{2x}-2+e^{-2x})$
Likewise.
$\cosh^2x-\sinh^2x=\tfrac14(4)=1$
Subtract: the exponentials cancel.

(\cosh t,\sinh t)

lies on the hyperbola

x^2-y^2=1

, hence the name.

TrickOsborn's rule Take any trig identity, swap to hyperbolic functions, and flip the sign of every product of two sines:

\cos2x=1-2\sin^2x

becomes

\cosh2x=1+2\sinh^2x

2. Inverse functions

\operatorname{arsinh}x=\ln(x+\sqrt{x^2+1})

$y=\operatorname{arsinh}x\Rightarrow x=\sinh y=\frac{e^y-e^{-y}}2$
Definition of the inverse.
$e^{2y}-2xe^y-1=0$
Multiply by $2e^y$ .
$e^y=x\pm\sqrt{x^2+1}$
Quadratic formula in $e^y$ .
$e^y=x+\sqrt{x^2+1}$
$x-\sqrt{x^2+1}<0$ , but $e^y>0$ .
$y=\ln\left(x+\sqrt{x^2+1}\right)$
Take logs.

\operatorname{arcosh}x=\ln\left(x+\sqrt{x^2-1}\right)\ (x\ge1),\qquad\operatorname{artanh}x=\tfrac12\ln\frac{1+x}{1-x}\ (|x|<1)

$\operatorname{arcosh}$: principal (non-negative) value.

3. Solving equations

Worked example $4\cosh x+\sinh x=4$

Solve exactly.

Goal: $x$ .
Strategy: Exponentials → quadratic in $e^x$ .

$2(e^x+e^{-x})+\tfrac12(e^x-e^{-x})=4$
Definitions.
$\tfrac52e^x+\tfrac32e^{-x}=4$
Collect.
$5e^{2x}-8e^x+3=0$
Multiply by $2e^x$ .
$(5e^x-3)(e^x-1)=0$
Factorise.

Answer

x=0

x=\ln\frac35

✓ Check

x=0

4(1)+0=4

✓.

Worked example $\cosh x=\frac54$

Solve.

Goal: $x$ .
Strategy: $\cosh$ is even: two solutions $\pm\operatorname{arcosh}\frac54$ .

$\operatorname{arcosh}\tfrac54=\ln\left(\tfrac54+\tfrac34\right)=\ln2$
Log form.
$x=\pm\ln2$
Even function.

Answer

x=\pm\ln2

✓ Check

\cosh\ln2=\frac{2+\frac12}2=\frac54

✓.

Common misconceptionForgetting $-\operatorname{arcosh}$

\cosh x=k>1

has two solutions;

\sinh x=k

has one.

Summary

Definitions in $e^x$ .
$\cosh^2x-\sinh^2x=1$ ; Osborn.
Inverses as logs.
Equations: quadratic in $e^x$ .

Catenary: Curve $y=a\cosh\frac xa$ of a hanging chain.
Osborn's rule: Converts trig identities to hyperbolic ones.

HistoryLambert and Riccati. Vincenzo Riccati and Johann Lambert developed hyperbolic functions in the 1760s, calling them "hyperbolic" because

(\cosh t,\sinh t)

traces the hyperbola

x^2-y^2=1

Fun factThe Gateway Arch. The 192 m St Louis arch is an inverted (weighted) catenary, a hyperbolic cosine curve, chosen so that the stresses run along the arch.

Section 3 of 14

Key formulae

Definitions

\sinh x=\frac{e^x-e^{-x}}2,\quad\cosh x=\frac{e^x+e^{-x}}2,\quad\tanh x=\frac{e^{2x}-1}{e^{2x}+1}

Key identity

\cosh^2x-\sinh^2x=1

Osborn's rule

\text{trig identity}\to\text{hyperbolic: change sign of any product of two sines}

Inverses

\operatorname{arsinh}x=\ln\!\left(x+\sqrt{x^2+1}\right),\ \operatorname{arcosh}x=\ln\!\left(x+\sqrt{x^2-1}\right),\ \operatorname{artanh}x=\tfrac12\ln\frac{1+x}{1-x}

TrickGo back to exponentials. When stuck, replace every

\sinh

and

\cosh

by its exponential form, multiply by

e^x

, and solve the quadratic in

e^x

Section 4 of 14

Worked examples

Example 1 · An equation0 / 3 steps

Solve

4\cosh x+\sinh x=4

Example 2 · arsinh as a log0 / 3 steps

Show

\operatorname{arsinh}x=\ln(x+\sqrt{x^2+1})

Section 5 of 14

Explore the graph

sinh and cosh from exponentials

Loading the interactive graph…

Try this

Both curves approach $\frac12e^x$ for large $x$ .
$\cosh$ is even (symmetric), $\sinh$ is odd, and $\cosh x>\sinh x$ always.

$\cosh x$ (U-shape) and $\sinh x$ (S-shape).

In the real worldSpecial relativity. Velocities add like

\tanh

: if

v=c\tanh\phi

, then combining two boosts just adds the "rapidities"

\phi

. That's why nothing exceeds the speed of light.

Section 6 of 14

Common mistakes

Incorrect

\cosh^2x+\sinh^2x=1

Correct

\cosh^2x-\sinh^2x=1

Osborn:

\sin^2

is a product of two sines, so its sign flips.

IncorrectTwo values for

\operatorname{arcosh}

Correct

\operatorname{arcosh}x\ge0

only

The principal value is non-negative;

\cosh x=k

has solutions

\pm\operatorname{arcosh}k

IncorrectKeeping

e^x<0

CorrectReject it

e^x

is always positive.

sin, cos, tan: we live on a circle

sinh, cosh, tanh: we live on a hyperbola and hang off bridges.

Section 7 of 14

Exam tip

Exam tipExact answers. Give logarithmic answers exactly, e.g.

\ln\frac35

, unless a decimal is asked for.

Section 8 of 14

Guided practice

SA-FP-HYP-G011 markL1 FoundationNon-calculatorDefinitions

Find

\cosh0

Salam Academy Original Practice · Proposed mark scheme

Section 9 of 14

Practice questions

SA-FP-HYP-0012 marksL2 StandardNon-calculatorDefinitions

Find the exact value of

\sinh(\ln2)

Salam Academy Original Practice · Proposed mark scheme

SA-FP-HYP-0022 marksL2 StandardNon-calculatorIdentities

Given

\sinh x=\frac34

, find

\tanh x

Salam Academy Original Practice · Proposed mark scheme

SA-FP-HYP-0033 marksL3 ExamCalculatorInverses and equations

Solve

\cosh x=\frac54

, giving both solutions to 4 s.f.

Salam Academy Original Practice · Proposed mark scheme

SA-FP-HYP-0042 marksL2 StandardNon-calculatorInverses and equations

Find the exact value of

e^{\operatorname{arsinh}(3/4)}

Salam Academy Original Practice · Proposed mark scheme

More questions on every skill in this subtopic are in the question bank under Practise.

Section 10 of 14

Challenge

SA-FP-HYP-0054 marksL4 ChallengingCalculatorInverses and equations

Solve

4\cosh x+\sinh x=4

. Give both solutions to 4 s.f.

Salam Academy Original Practice · Proposed mark scheme

Section 11 of 14

Summary

Definitions in $e^x$ .
$\cosh^2x-\sinh^2x=1$ ; Osborn's rule.
Inverses as logarithms.
Equations: quadratic in $e^x$ .

Section 12 of 14

Quick quiz

Q1 · SA-FP-HYP-Q011 markL1 FoundationNon-calculatorIdentities

Which identity is correct?

Salam Academy Original Practice · Proposed mark scheme

Section 13 of 14

Same topic, different board

Specification: 7367 A-level Further Mathematics
Where: Paper 1 and 2 · Hyperbolic functions
Expected: Definitions, graphs, identities, inverses and solving equations.
Calculator: Permitted on all papers.
Typical style: AQA mixes short multiple-choice questions with longer structured ones.
Open in: 7367 · Paper 1 7367 · Paper 2

Section 14 of 14

Past paper questions

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Salam Academy original content · Content status: reviewed and published. Syllabus mapping: AQA 7367, First teaching 2017 (official specification).